For a curve of equation 2ye^-3x -x = 4, find dy/dx

here the student needs to use both implicit differentiation and the product rule.

I would differentiate term by term for this problem.

for 2ye^-3x you have to use the proudct rule. uv differentiates to uv' +u'v 

so the above differentiates to -6ye^-3x +2(dy/dx)e^-3x

-x differentiates to -1

4 is a constant so differentiates to 0.

leads to ; -6ye^-3x +2(dy/dx)e^-3x -1=0

which can be written as 2(dy/dx)e^-3x = 6ye^-3x +1

leads to dy/dx= (6ye^-3x +1)/2e^-3x =(e^3x)/2 + 3y